Trapezoidal Rule Calculator for a Function

Approximate the integral $$$\int\limits_{0}^{1} \sqrt{\sin^{3}{\left(x \right)} + 1}\, dx$$$ with $$$n = 5$$$ using the trapezoidal rule.

The trapezoidal rule uses trapezoids to approximate the area:

$$$\int\limits_{a}^{b} f{\left(x \right)}\, dx\approx \frac{\Delta x}{2} \left(f{\left(x_{0} \right)} + 2 f{\left(x_{1} \right)} + 2 f{\left(x_{2} \right)} + 2 f{\left(x_{3} \right)}+\dots+2 f{\left(x_{n-2} \right)} + 2 f{\left(x_{n-1} \right)} + f{\left(x_{n} \right)}\right)$$$

where $$$\Delta x = \frac{b - a}{n}$$$.

We have that $$$f{\left(x \right)} = \sqrt{\sin^{3}{\left(x \right)} + 1}$$$, $$$a = 0$$$, $$$b = 1$$$, and $$$n = 5$$$.

Therefore, $$$\Delta x = \frac{1 - 0}{5} = \frac{1}{5}$$$.

Divide the interval $$$\left[0, 1\right]$$$ into $$$n = 5$$$ subintervals of the length $$$\Delta x = \frac{1}{5}$$$ with the following endpoints: $$$a = 0$$$, $$$\frac{1}{5}$$$, $$$\frac{2}{5}$$$, $$$\frac{3}{5}$$$, $$$\frac{4}{5}$$$, $$$1 = b$$$.

Now, just evaluate the function at these endpoints.

$$$f{\left(x_{0} \right)} = f{\left(0 \right)} = 1$$$

$$$2 f{\left(x_{1} \right)} = 2 f{\left(\frac{1}{5} \right)} = 2 \sqrt{\sin^{3}{\left(\frac{1}{5} \right)} + 1}\approx 2.007826067912793$$$

$$$2 f{\left(x_{2} \right)} = 2 f{\left(\frac{2}{5} \right)} = 2 \sqrt{\sin^{3}{\left(\frac{2}{5} \right)} + 1}\approx 2.058206972332648$$$

$$$2 f{\left(x_{3} \right)} = 2 f{\left(\frac{3}{5} \right)} = 2 \sqrt{\sin^{3}{\left(\frac{3}{5} \right)} + 1}\approx 2.17257446116512$$$

$$$2 f{\left(x_{4} \right)} = 2 f{\left(\frac{4}{5} \right)} = 2 \sqrt{\sin^{3}{\left(\frac{4}{5} \right)} + 1}\approx 2.340214753424868$$$

$$$f{\left(x_{5} \right)} = f{\left(1 \right)} = \sqrt{\sin^{3}{\left(1 \right)} + 1}\approx 1.263258974474734$$$

Finally, just sum up the above values and multiply by $$$\frac{\Delta x}{2} = \frac{1}{10}$$$: $$$\frac{1}{10} \left(1 + 2.007826067912793 + 2.058206972332648 + 2.17257446116512 + 2.340214753424868 + 1.263258974474734\right) = 1.084208122931016.$$$

$$$\int\limits_{0}^{1} \sqrt{\sin^{3}{\left(x \right)} + 1}\, dx\approx 1.084208122931016$$$A